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It's been a while since I used residue theorem to evaluate anything. I remember that whenever we have a real valued function, we can use residue theorem to evaluate its integral with associated complex integral etc.

Anyway, I'd have to evaluate the following, already complex-valued intergal: $$\int \limits_{-\infty}^\infty \dfrac{\mathrm{e}^{-ixt}}{a^2 - x^2 - ibx}\mathrm{d}x$$

Here $a$, $b$, $t$ and $x$ are real. Edit: actually, also $t>0$ and $b>0$. Ok, I can find the poles of the function and evaluate the residues around them. I've been given a hint to use the residue theorem, but how do I apply it here? I think I could apply it if I was able to visualise the given integration path, "complete it" with another path (so that this path wouldn't contribute in the end).

But how to choose it? I have trouble visualising the given integration path, what does it look like?

Thanks a lot in advance!

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Because $t \gt 0$, the contour needs to be a semicircle of radius $R$ in the lower half-plane - not in the upper half plane. That is, consider the contour integral

$$\oint_C dz \frac{e^{-i z t}}{a^2-z^2-i b z} $$

where $C$ is the above-described semicircle in the lower half-plane, positively oriented. Note that the poles of the integrand are both contained within $C$:

$$z_{\pm} = -i \frac{b}{2} \pm i \frac12 \sqrt{b^2-4 a^2}$$

The contour integral, positively oriented (i.e., traversed counterclockwise), is equal to

$$-\int_{-R}^R dx \frac{e^{-i x t}}{-x^2-i b x+a^2} + i R \int_{-\pi}^0 d\theta \, e^{i \theta} \frac{e^{-i t R e^{i \theta}}}{a^2-R^2 e^{i 2 \theta}-i b R e^{i \theta}} $$

We can show that the second integral vanishes as $R\to\infty$ by noting that the magnitude of the integral is bounded by

$$\frac1{R} \int_{-\pi}^0 d\theta \, e^{t R \sin{\theta}} \le \frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{t R^2}$$

The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles within the contour $C$. As both poles are contained within $C$, we have that

$$\int_{-\infty}^{\infty} dx \frac{e^{-i x t}}{-x^2-i b x+a^2} = \begin{cases}\displaystyle \frac{2 \pi}{\sqrt{b^2-4 a^2}} e^{-b t/2} \sin{\left (\sqrt{b^2-4 a^2}\, t \right )} & b \gt 2 a\\ \displaystyle\frac{2 \pi}{\sqrt{4 a^2-b^2}} e^{-b t/2} \sinh{\left (\sqrt{4 a^2-b^2}\, t \right )} & b \lt 2 a \\ 2 \pi t \, e^{-b t/2} & b=2 a\end{cases} $$

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  • $\begingroup$ Wow, thank you! So applying residue theorem works just the same way, even though the original integral isn't real valued? $\endgroup$ – souf Apr 9 '14 at 10:08
  • $\begingroup$ @souf: yep ${}{}{}$ $\endgroup$ – Ron Gordon Apr 9 '14 at 10:13
  • $\begingroup$ Ok! I think i got it now: is this because the original integration variable is real? [So from -infty to infty we just have one possible path to integrate x, since x is real?] $\endgroup$ – souf Apr 9 '14 at 10:20
  • $\begingroup$ @souf: the idea is simply to define the integration contour such that one piece gives you the integral you seek, while the others either vanish or are easily evaluated. When evaluating real integrals, then yes, we go along the real axis. $\endgroup$ – Ron Gordon Apr 9 '14 at 10:21
  • $\begingroup$ Yes. Does "real integral" here mean that the variable of integration is real, or that the function only gets real values? $\endgroup$ – souf Apr 9 '14 at 10:24

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